मराठी

Evaluate the Following Integral: 2 ∫ − 2 | 2 X + 3 | D X - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]
बेरीज

उत्तर

\[\int_{- 2}^2 \left| 2x + 3 \right| d x\]
\[We\ know\ that\ \left| 2x + 3 \right| = \begin{cases} - \left( 2x + 3 \right) &, &- 2 \leq x \leq - \frac{3}{2}\\\left( 2x + 3 \right)&, &- \frac{3}{2} < x \leq 2\end{cases}\]
\[ \therefore I = \int_{- 2}^\frac{- 3}{2} - \left( 2x + 3 \right) d x + \int_{- \frac{3}{2}}^2 \left( 2x + 3 \right) d x\]
\[ \Rightarrow I = - \left[ x^2 + 3x \right]_{- 2}^\frac{- 3}{2} + \left[ x^2 + 3x \right]_{- \frac{3}{2}}^2 \]
\[ \Rightarrow I = - \frac{9}{4} + \frac{9}{2} + 4 - 6 + 4 + 6 - \frac{9}{4} + \frac{9}{2}\]
\[ \Rightarrow I = \frac{25}{2}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.3 [पृष्ठ ५६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.3 | Q 5 | पृष्ठ ५६

संबंधित प्रश्‍न

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

\[\int\limits_1^2 \left( \frac{x - 1}{x^2} \right) e^x dx\]

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

\[\int\limits_0^1 x e^{x^2} dx\]

\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

\[\int_0^\frac{\pi}{2} \frac{\cos x}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^n}dx\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

\[\int\limits_0^{\pi/2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} dx\]

\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

\[\int\limits_{- 1}^1 \log\left( \frac{2 - x}{2 + x} \right) dx\]

\[\int\limits_1^3 \left( 2x + 3 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 4 \right) dx\]

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_a^b x\ dx\]

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_0^{\pi/2} \cos^2 x\ dx .\]

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]

 


If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


\[\int\limits_0^2 \left[ x \right] dx .\]

The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 


\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to


\[\int\limits_0^\pi x \sin x \cos^4 x dx\]


\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]


\[\int\limits_0^4 x dx\]


\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Choose the correct alternative:

`Γ(3/2)`


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×